Saved Bookmarks
| 1. |
For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculated from their molecular formula.The observed densities decreases towards a limit as the temperature is raised.This is due to the splitting of the molecular into simpler ones.The process is reversible and is called thermal dissociation. Example : `NH_4Cl hArr NH_3+HCl I_2hArr 2I` `N_2O_4 hArr 2NO_2 PCl_5hArr PCl_3+Cl_2` With increase in the number of molecules, the volume increases (pressure remaining constant) and in consequence, the density decreases.As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occurs the density reaches its lowest limit. The extent of dissociation, i.e., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation.Gas density measurements can be used to determine the degree of dissociation.Let us take by general case where one molecule of a substance A splits up into n molecule of A on heating , i.e., `A_n(g)hArr nA(g)` `t=0 " " a" " 0` `t=t_(eq) a-x n.x alpha=x/a rArr x=a alpha.` `a-a alpha naalpha` Total no. of moles =`a-a alpha+n a alpha` =`[1+(n-1)alpha]a` Observed molecular weight or molar mass of the mixture `M_("mixture")=M_(A_(n))/([1+(n-1)alpha]), M_(A_(n))`=Molar mass of gas `A_n` The `K_P` for the reaction `N_2O_4 hArr 2NO_2` is 640 mm at 775 K.The percentage dissociation of `N_2O_4` at equilibrium pressure of 160 mm is :A. `80%`B. `30%`C. `50%`D. `70%` |
|
Answer» Correct Answer - D `{:(,N_2O_4" "hArr,2NO_2),("Mole before equilibrium",1,),("Mole at equilibrium",(1-x),2x):}` `K_p=(4x^2)/((1-x))xx[P/(sumn)]^(Deltan)implies 640=(4x^2)/((1-x))160/((1+x))` `4=(4x^2)/((1-x))` or `1-x^2=x^2` or `2x^2=1` `:. " " x^2=1//2 or x=0.707=70.7%` |
|